Simplify; express your answer in exponential form. Assume $t\neq 0, z\neq 0$. $\dfrac{{(t^{2})^{-5}}}{{(t^{-4}z^{-5})^{-3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${t^{2}}$ to the exponent ${-5}$ . Now ${2 \times -5 = -10}$ , so ${(t^{2})^{-5} = t^{-10}}$ In the denominator, we can use the distributive property of exponents. ${(t^{-4}z^{-5})^{-3} = (t^{-4})^{-3}(z^{-5})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(t^{2})^{-5}}}{{(t^{-4}z^{-5})^{-3}}} = \dfrac{{t^{-10}}}{{t^{12}z^{15}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{-10}}}{{t^{12}z^{15}}} = \dfrac{{t^{-10}}}{{t^{12}}} \cdot \dfrac{{1}}{{z^{15}}} = t^{{-10} - {12}} \cdot z^{- {15}} = t^{-22}z^{-15}$.